POJ 2002 Squares

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14440		Accepted: 5441

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property. 

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates. 

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1

分析：hash。

Code：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#define eps 1e-7
#define LL long long
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;

const int MOD=40000;
struct node{
    int x,y;
}p[1005];
int Hash[MOD+MOD],next[1005];
int n;

bool cmp(node a,node b){
    if(a.x==b.x) return a.y>b.y;
    return a.x<b.x;
}

bool find(int x,int y){
    int tmp=x+y;
    if(tmp<0) tmp+=MOD;
    for(int i=Hash[tmp];i!=-1;i=next[i]){
        if(p[i].x==x&&p[i].y==y) return true;
    }
    return false;
}

int main()
{
    while(scanf("%d",&n),n){
        for(int i=0;i<n;i++) scanf("%d %d",&p[i].x,&p[i].y);
        sort(p,p+n,cmp);
        memset(Hash,-1,sizeof(Hash));
        for(int i=0;i<n;i++){
            int tmp=p[i].x+p[i].y;
            if(tmp<0) tmp+=MOD;
            next[i]=Hash[tmp];
            Hash[tmp]=i;
        }
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                int a=p[i].x+p[i].y-p[j].y;
                int b=p[i].y+p[j].x-p[i].x;
                if(!find(a,b)) continue;
                a=p[j].x+p[i].y-p[j].y;
                b=p[j].y+p[j].x-p[i].x;
                if(!find(a,b)) continue;
                ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}